Question

A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the light leave the water?

Answer 1

Answer:

45.93°

Explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is $n_1=1.33$

Refractive index of the air that is $n_2=1$ (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law $n_1sini=n_2sinr$

$1.33sin32.7=1\times sinr$

$sinr=0.7185$

r =45.93°

So the light leave the water at an angle of 45.93°